Download presentation
Presentation is loading. Please wait.
Published byVeronika Tanudjaja Modified over 5 years ago
1
Yan Huang - CSCI5330 Database Implementation – Query Processing
1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
2
Yan Huang - CSCI5330 Database Implementation – Query Processing
Exercise Compute depositor customer, with depositor as the outer relation. Customer has a secondary B+-tree index on customer-name Blocking factor 20 keys #customer = 400b/10,000t #depositor =100b/5,000t Merge join log(400) + log(100) = 516 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
3
Yan Huang - CSCI5330 Database Implementation – Query Processing
Hybrid Merge-join If one relation is sorted, and the other has a secondary B+-tree index on the join attribute Merge the sorted relation with the leaf entries of the B+-tree . Sort the result on the addresses of the unsorted relation’s tuples Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples Sequential scan more efficient than random lookup 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
4
Yan Huang - CSCI5330 Database Implementation – Query Processing
Hybrid Merge-join r … s … … 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
5
Yan Huang - CSCI5330 Database Implementation – Query Processing
Hash join Hash function h, range 0 k Buckets for R1: G0, G1, ... Gk Buckets for R2: H0, H1, ... Hk Algorithm (1) Hash R1 tuples into G buckets (2) Hash R2 tuples into H buckets (3) For i = 0 to k do match tuples in Gi, Hi buckets 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
6
Simple example hash: even/odd
R1 R2 Buckets 2 5 Even R R2 Odd: 8 13 11 14 2 4 8 3 5 9 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
7
Yan Huang - CSCI5330 Database Implementation – Query Processing
Example Hash Join R1, R2 contiguous (un-ordered) Use 100 buckets Read R1, hash, + write buckets R1 100 ... ... 10 blocks 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
8
Then repeat for all buckets
-> Same for R2 -> Read one R1 bucket; build memory hash table -> Read corresponding R2 bucket + hash probe R1 Relation R1 is called the build input and R2 is called the probe input. R2 ... R1 ... memory Then repeat for all buckets 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
9
Yan Huang - CSCI5330 Database Implementation – Query Processing
Cost: “Bucketize:” Read R1 + write Read R2 + write Join: Read R1, R2 Total cost = 3 x [ bR1+bR2] Note: this is an approximation since buckets will vary in size and we have to round up to blocks 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
10
Minimum memory requirements:
Size of R1 bucket =(x/k) k = number of memory buffers x = number of R1 blocks So... (x/k) < k k > x Which relation should be the build relation? 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
11
Example of Cost of Hash-Join
customer depositor Assume that memory size is 20 blocks bdepositor= 100 and bcustomer = 400. depositor is to be used as build input. Partition it into five partitions, each of size 20 blocks. This partitioning can be done in one pass. Similarly, partition customer into five partitions,each of size 80. This is also done in one pass. Therefore total cost: 3( ) = 1500 block transfers ignores cost of writing partially filled blocks 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
12
Yan Huang - CSCI5330 Database Implementation – Query Processing
Complex Joins Join with a conjunctive condition: r 1 2... n s Either use nested loops/block nested loops, or Compute the result of one of the simpler joins r i s final result comprises those tuples in the intermediate result that satisfy the remaining conditions 1 i –1 i +1 n Join with a disjunctive condition r 1 2 ... n s Compute as the union of the records in individual joins r i s: (r 1 s) (r 2 s) (r n s) 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
13
Yan Huang - CSCI5330 Database Implementation – Query Processing
Other Operations Duplicate elimination can be implemented via hashing or sorting. Projection is implemented by performing projection on each tuple followed by duplicate elimination. 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
14
Other Operations : Aggregation
Sorting Hashing 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
15
Other Operations : Set Operations
Set operations (, and ) can either use variant of merge-join after sorting or variant of hash-join. 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
16
Evaluation of Expressions
So far: we have seen algorithms for individual operations Alternatives for evaluating an entire expression tree Materialization: generate results of an expression whose inputs are relations or are already computed, materialize (store) it on disk. Repeat. Pipelining: pass on tuples to parent operations even as an operation is being executed 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
17
Yan Huang - CSCI5330 Database Implementation – Query Processing
Q Query Plan Focus: Relational System Others? 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
18
Yan Huang - CSCI5330 Database Implementation – Query Processing
Example Select B,D From R,S Where R.A = “c” S.E = 2 R.C=S.C 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
19
Relational Algebra - can be used to describe plans...
Ex: Plan I B,D sR.A=“c” S.E=2 R.C=S.C X R S OR: B,D [ sR.A=“c” S.E=2 R.C = S.C (RXS)] 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
20
Yan Huang - CSCI5330 Database Implementation – Query Processing
Another idea: B,D sR.A = “c” sS.E = 2 R S Plan II natural join 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
21
Example: Estimate costs
L.Q.P P1 P2 …. Pn C1 C2 …. Cn Pick best! 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
22
Relational algebra optimization
Transformation rules (preserve equivalence) What are good transformations? 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
23
Rules: Natural joins & cross products & union
R S = S R (R S) T = R (S T) 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
24
Yan Huang - CSCI5330 Database Implementation – Query Processing
Note: Carry attribute names in results, so order is not important Can also write as trees, e.g.: T R R S S T 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
25
Rules: Natural joins & cross products & union
R S = S R (R S) T = R (S T) R x S = S x R (R x S) x T = R x (S x T) R U S = S U R R U (S U T) = (R U S) U T 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
26
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules: Selects sp1p2(R) = sp1vp2(R) = sp1 [ sp2 (R)] [ sp1 (R)] U [ sp2 (R)] 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
27
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules: Project Let: X = set of attributes Y = set of attributes XY = X U Y pxy (R) = px [py (R)] 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
28
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules: s combined Let p = predicate with only R attribs q = predicate with only S attribs m = predicate with only R,S attribs sp (R S) = sq (R S) = [sp (R)] S R [sq (S)] 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
29
Rules: s + combined (continued)
Some Rules can be Derived: spq (R S) = spqm (R S) = spvq (R S) = 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
30
Yan Huang - CSCI5330 Database Implementation – Query Processing
spq (R S) = [sp (R)] [sq (S)] spqm (R S) = sm [(sp R) (sq S)] spvq (R S) = [(sp R) S] U [R (sq S)] 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
31
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules: p,s combined Let x = subset of R attributes z = attributes in predicate P (subset of R attributes) px[sp (R) ] = px pxz {sp [ px (R) ]} 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
32
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules: p, combined Let x = subset of R attributes y = subset of S attributes z = intersection of R,S attributes pxy (R S) = pxy{[pxz (R) ] [pyz (S) ]} 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
33
Yan Huang - CSCI5330 Database Implementation – Query Processing
pxy {sp (R S)} = pxy {sp [pxz’ (R) pyz’ (S)]} z’ = z U {attributes used in P } 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
34
Rules for s, p combined with X
similar... e.g., sp (R X S) = ? 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
35
Yan Huang - CSCI5330 Database Implementation – Query Processing
Rules s, U combined: sp(R U S) = sp(R) U sp(S) sp(R - S) = sp(R) - S = sp(R) - sp(S) 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
36
Which are “good” transformations?
sp1p2 (R) sp1 [sp2 (R)] sp (R S) [sp (R)] S R S S R px [sp (R)] px {sp [pxz (R)]} 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
37
Conventional wisdom: do projects early
Example: R(A,B,C,D,E) x={E} P: (A=3) (B=“cat”) px {sp (R)} vs. pE {sp{pABE(R)}} 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
38
What if we have A, B indexes?
But B = “cat” A=3 Intersect pointers to get pointers to matching tuples 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
39
Yan Huang - CSCI5330 Database Implementation – Query Processing
Bottom line: No transformation is always good Usually good: early selections 1/14/2005 Yan Huang - CSCI5330 Database Implementation – Query Processing
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.